If $D\in [AB$ s.t. $AD=BC$ and $\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$ find $\angle {A}$. Let $\triangle {ABC}$ s.t. $AB=AC$ and $\angle{A}>90$. If $D\in [AB$ s.t. $AD=BC$ and  $\angle{ADC}=\frac{3}{4}\cdot \angle{ABC}$ find $\angle {A}$.
My idea: I denote $\angle B=4x$, then I apply "Sine theorem" in $\triangle {ABC}$ and $ \triangle {ADC}$. 
I obtain $sin(5x)=2sin(3x)\cdot cos(4x)$. Now I am stuck.
I try to construct and to solve it with an elementary construction but I didn't succeed. Can I apply here "The pants theorem"?
 A: 
Let $E$ on $BC$ such that $CE\cong AC\cong AB$, and denote $\angle ABC = \alpha$.


*

*$\triangle BDE$ is isosceles, therefore $\angle BDE =\angle BED= \frac{\alpha}2$.

*Since $\angle ADC = \frac34\alpha$, we have $\angle EDC = \frac14 \alpha$.

*Note that $\angle DEC = 180^\circ -\frac12\alpha$, so that $\triangle DEC$ is isoceles, too.

*Construct then the rhombus $DGCE$ and connect $A$ with $G$. Observe that $\triangle AGC$ is equilateral, $\triangle ADG$ is isosceles, and $DG\parallel BC$.

*We can write therefore $\angle BAC$ in two ways and obtain the equation $$180^\circ - 2\alpha = 60^\circ + \alpha.$$

*Thus $\alpha = 40^\circ$ and $\angle BAC = 100^\circ$.

A: Continue with what you have and apply the trig identity $2\cos a\sin b = \sin(a+b)-\sin(a-b)$,
$$\sin5x=2\cos4x\sin3x=\sin7x-\sin x$$
Rearrange and apply the above identity one more time,
$$\sin x = \sin7x - \sin5x = 2\cos 6x\sin x$$
which yields $\cos6x = \frac12$. Thus, $x=10^\circ$, $\angle B = 40^\circ$ and $\angle A = 180^\circ - 2\angle B = 100^\circ$.
